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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:

Convex quadrilateral $ABCD$ has right angles $\angle A$ and $\angle C$ and is such that $AB = BC$ and $AD = CD$ . The diagonals $AC$ and $BD$ intersect at point $M$ . Points $P$ and $Q$ lie on the circumcircle of triangle $AMB$ and segment $CD$ , respectively, such that points $P$ , $M$ , and $Q$ are collinear. Suppose that $m \angle ABC = 160^\circ$ and $m \angle QMC = 40^\circ$ . Find $MP \cdot MQ$ , given that $MC = 6$ .

Solution

Solution:

Answer: $36$ . Note that $m \angle QPB = m \angle MPB = m \angle MAB = m \angle CAB = \angle BCA = \angle CDB$ . Thus, $MP \cdot MQ = MB \cdot MD$ . On the other hand, segment $CM$ is an altitude of right triangle $BCD$ , so $MB \cdot MD = MC^{2} = 36$ .

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